The orbit-degree decomposition conjecture for finite-field algebras

Theorem. Let $q$ be any prime power and let

$$f(T)=a_0+a_1T+\cdots+a_dT^d\in\mathbf F_q[T], \qquad d\ge1,\quad a_0a_d\ne0.$$

Put

$$R_f=\mathbf F_q[T]/(f),\qquad L_f(X)=\sum_{i=0}^d a_iX^{q^i},\qquad S_f=\mathbf F_q[X]/(L_f).$$

If $\mathcal O$ ranges over the orbits of multiplication by $T$ on the underlying set of $R_f$, then

$$S_f\cong\prod_{\mathcal O}\mathbf F_{q^{|\mathcal O|}}.$$

Proof. Because $a_0\ne0$, the class of $T$ is a unit in the finite ring $R_f$. Let $m$ be its multiplicative order. Then

$$T^m-1=f(T)h(T)$$

for some $h\in\mathbf F_q[T]$. Write $F(u)=u^q$ for Frobenius on $K=\mathbf F_{q^m}$. The roots of $L_f$ are exactly $V=\ker f(F)$. Every such root belongs to $K$, since

$$(F^m-I)u=h(F)f(F)u=0.$$

The normal-basis theorem gives a basis

$$\theta,F\theta,\ldots,F^{m-1}\theta$$

of $K$ over $\mathbf F_q$. Consequently,

$$\mathbf F_q[T]/(T^m-1)\longrightarrow K, \qquad[g]\longmapsto g(F)\theta,$$

is an isomorphism of $\mathbf F_q[T]$-modules. Therefore

$$V\cong\ker\left(f:\mathbf F_q[T]/(fh)\to\mathbf F_q[T]/(fh)\right) =h\mathbf F_q[T]/(fh).$$

Multiplication by $h$ induces

$$\mathbf F_q[T]/(f)\xrightarrow{\sim}h\mathbf F_q[T]/(fh), \qquad[g]\longmapsto[gh].$$

Indeed, $fh\mid gh$ implies $f\mid g$, so this map is injective and plainly surjective. No coprimality between $f$ and $h$ is needed.

Thus $V\cong R_f$ as $\mathbf F_q[T]$-modules, with $T$ acting as Frobenius on $V$ and as multiplication by $T$ on $R_f$. Their orbit multisets therefore coincide. Since

$$L_f'(X)=a_0\ne0,$$

the polynomial $L_f$ is separable. Its irreducible factors correspond exactly to Frobenius orbits of roots, and their degrees equal the corresponding orbit lengths. The Chinese remainder theorem now gives

$$S_f\cong\prod_{\mathcal O}\mathbf F_{q^{|\mathcal O|}}.$$

Taking $q=2$ and $f(T)=T^n+T+1$ yields

$$L_f(X)=X^{2^n}+X^2+X$$

and proves the conjectured decomposition

$$\boxed{\displaystyle S_n\cong\prod_{\mathcal O}\mathbf F_{2^{|\mathcal O|}}}$$

for every $n\ge2$, including the zero orbit. This orbit-degree result makes no assertion about multiplicative primitivity of individual roots.